EE2257 CS LAB

MAILAM ENGINEERING COLLEGEMAILAM – 604 304

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

EE 2257 – CONTROL SYSTEM LABORATORY MANUAL

NAME :

ROLL NO. :

CLASS : B.E. / EEE - B

YEAR / SEM : II Yr / IV Sem

SYLLABUS

EE 2257 – CONTROL SYSTEM LABORATORY

LIST OF EXPERIMENTS

1. Determination of transfer function of DC Servomotor

2. Determination of transfer function of AC Servomotor

3. Analog simulation of Type - 0 and Type – 1 systems

4. Determination of transfer function of DC Generator

5. Determination of transfer function of DC Motor

6. Stability analysis of linear systems

7. DC and AC position control systems

8. Stepper motor control system

9. Digital simulation of first order systems

10. Digital simulation of second order systems

CYCLE - 1

LIST OF EXPERIMENTS

1. (a)Determination of transfer function of armature controlled DC Servomotor. (b)Determination of transfer function of field controlled DC Servomotor.

2. Determination of transfer function of AC Servomotor.

3. Analog simulation of Type - 0 and Type – 1 systems.

4. Determination of transfer function of DC Generator.

5. (a) Determination of transfer function of armature controlled DC Motor. (b) Determination of transfer function of field controlled DC Motor.

6. Digital simulation of first order systems.

CYCLE - 2

LIST OF EXPERIMENTS

7. Stability analysis of linear systems.

8. DC and AC position control systems.

9. Stepper motor control system.

10. Digital simulation of second order systems.

INDEX

EX.NO. DATE NAME OF THE EXPERIMENT MARKS SIGNATURE

Ex. No. – 1(a)

DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF ARMATURE CONTROLLED DC SERVOMOTOR

Aim:To determine the transfer function parameters of armature controlled DC servomotor.

Apparatus Required:

S.No. Apparatus Quantity1. DC servomotor trainer kit 1

2. DC servomotor 1

Name plate details of DC servomotor:

Armature voltage 48 V DCField voltage 48 V DCArmature current 2.5 APower 100 WSpeed 1500 rpm

Theory:

DC Servo motor is basically a torque transducer which converts electrical energy into mechanical energy It is basically a separately excited type DC motor. The torque developed on the motor shaft is directly proportional to the field flux and armature current, Tm = Km Φ Ia. The back emf developed by the motor is Eb = Kb Φ ωm..

Fig.1.1 Circuit diagram of armature controlled DC servomotor.

Front Panel diagram:

In an armature controlled DC Servo motor, the control signal available from the servo amplifier is applied to the armature of the motor. This signal is based on the

feedback information, supplied to the controller. Due to this armature current changes which in turn changes the torque produced. The field winding is supplied with constant current hence the flux remains constant. Therefore these motors are called as constant

magnetic flux motors.

Derivation of transfer function:Let, Ra = Armature Resistance La = Armature Inductance Eb = Back emf Ia = Armature current T = Torque developed in the motor J = Moment of inertia B = Dashpot

= Angular DisplacementKt = Armature torque constantKb = Back emf constant

From equivalent circuit, Differential equation of electrical circuit can be written as Ra ia +La dia/dt + eb = Va ------------------------ 1

Differential equation of mechanical system can be written as

J d2 /dt2 + B d / dt = T ------------------------ 2

Torque is directly proportional to the armature current T Ia , T Ia T = Kt Ia ------------------------ 3

Motor back emf is directly proportional to the speed (Angular velocity ) Eb = Kb d / dt ------------------------ 4

Taking laplace transform for equations 1, 2, 3 & 4 we get

Ra Ia(s) + Las Ia(s) + Eb(s) = Va(s) (Ra + Las) Ia(s) + Eb(s) = Va(s) ------------------------ 5

Js2(s) + Bs(s) = T(s) ------------------------ 6

T(s) = Kt Ia(s) ------------------------ 7

Eb(s) = Kbs (s) ------------------------ 8

Equating equation 6 & 7 we get, KtIa(s) = Js2(s) + Bs(s)

KtIa(s) = (Js2 + Bs) (s)

Ia(s) = (Js2 + Bs) (s) / Kt ------------------------ 9

Substitute eqn. 8 & 9 in 5,

[ (Ra + Las) (Js2 + Bs) (s) / Kt ] + Kbs (s) = Va(s)

(s) (Ra + Las) (Js2 + Bs) + Kb Kt s / Kt = Va(s)

(s) Kt = Va(s) (Ra + Las) (Js2 + Bs) + Kb Kt s

Procedure:

(a) To find Ra:1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the armature terminals of the trainer kit to the motor armature terminal.5. Switch ON the power switch and SPST switch S1.6. Shaft should not rotate, when applying DC input voltage.7. Note the armature voltage and current for the various values of armature voltage. 8. Calculate armature resistance Ra.

(b) To find La:1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the AC voltage terminal of the trainer kit to the motor armature terminal.5. Switch ON the power.6. Shaft should not rotate, when applying AC input voltage.7. Note the AC voltage and current for the various values of AC voltage. 8. Calculate armature inductance La.

(c) To find Kb:

1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the armature terminals of the trainer kit to the motor armature terminal.5. Connect the field terminals of the trainer kit to the motor field terminal.6. Switch ON the power switch and SPST switches S1 and S2.7. Set the field voltage at rated value (48 V).8. By varying the armature voltage, note the corresponding armature current and

speed till the rated speed is reached.

9. Plot the graph Eb versus ω. From the graph calculate the value of Kb.

(d) To find Kt:1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the armature terminals of the trainer kit to the motor armature terminal.5. Connect the field terminals of the trainer kit to the motor field terminal.6. Switch ON the power switch and SPST switches S1 and S2.7. Set the field voltage at rated value (48 V).8. Vary armature voltage till motor runs at rated speed.9. By varying the load, note the armature voltage, current and spring balance

readings.10. Plot the graph T versus armature current. From the graph calculate the value of Kt.

Tabulation:

To find Ra :

Average Ra = _________ Ω

To find La :

Armature voltage

( Va)

Armature current

( Ia)

Armature Impedence Za=Va/Ia

Armature ReactanceXa=:( Za2 - Ra

2)

Armature InductanceLa= Xa/ 2 f

Average La = _________ H

To find Kb :

Armature voltage( Va)

Armature current( Ia)

Armature ResistanceRa = Va / Ia

Armature voltage

( Va)

Armature current

( Ia)

Speed( N )Rpm

Eb =Va - IaRa

Volts =2N/60

rad/sec

To find Kt :

Armature current

( Ia)

Armature voltage

( Va)

S1

(Kg)S2

(Kg)S1 - S2

(Kg)Torque

(Ta) = 9.81(S1-S2) R Nm

Model Graph:

Measurement of Back Emf constant (Kb) :

Eb

Eb

(radians/sec)

Measurement of Kt :

Torque Vs Armature current (Ia)

T(Nm) T

Ia

Ia (A)

Model calculation :

Kb = Eb /

Kt = T/ Ia

B = 0.001 Nm/ (rad/sec)J =0.0074 Kg m2

R = 0.075 m

Result:Thus the transfer function of the armature controlled DC servomotor is determined as

Ex. No. – 1(b)

DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF FIELD CONTROLLED DC SERVOMOTOR

Aim:To determine the transfer function parameters of field controlled DC servomotor.

Apparatus Required:

S.No. Apparatus Quantity1. DC servomotor trainer kit 1

2. DC servomotor 1

Name plate details of DC servomotor:

Armature voltage 48 V DCField voltage 48 V DCArmature current 2.5 APower 100 WSpeed 1500 rpm

Theory:

DC Servo motor is basically a torque transducer which converts electrical energy into mechanical energy. It is basically a separately excited type DC motor. The torque developed on the motor shaft is directly proportional to the field flux and armature current, Tm = Km Ia. The back emf developed by the motor is Eb = Kb m.

In a field controlled DC Servo motor, the electrical signal is externally applied to the field winding. Hence current through field winding is controlled in turn controlling the flux. In a control system, a controller generates the error signal by comparing the actual o/p with the reference i/p. Such an error signal is not enough to drive the DC motor. Hence it is amplified by the servo amplifier and applied to the field winding. With the help of constant current source, the armature current is maintained constant.

When there is change in voltage applied to the field winding, the current through the field winding changes. This changes the flux produced by field winding. This motor has large Lf / Rf ratio, so time constant of this motor is high and it can’t give rapid responses to the quick changing control signals.

Front panel diagram:

Fig. 1.2 Circuit diagram of field controlled DC servomotor.

Derivation of transfer function:Let,

Rf = field Resistance Lf = field Inductance T = Torque developed in the motor J = Moment of inertia B = Dashpot

= Angular DisplacementKf = Field torque constant

From equivalent circuit, Differential equation of electrical circuit can be written as Rf if +Lf dif/dt = Vf ------------------------ 1

Differential equation of mechanical system can be written as

J d2 /dt2 + B d / dt = T ------------------------ 2

Torque is directly proportional to the armature current T Ia , T If T = Kf If ------------------------ 3

Taking laplace transform for equations 1, 2, & 3 we get

Rf If(s) + s Lf If(s) = Vf(s) (Rf + s Lf) If(s) = Vf(s) ------------------------ 4

Js2(s) + Bs(s) = T(s) ------------------------ 5

T(s) = Kf If(s) ------------------------ 6

Equating equation 5 & 6 we get, Kf If(s) = Js2(s) + Bs(s)

Kf If(s) = (Js2 + Bs) (s)

If(s) = (Js2 + Bs) (s) / Kf ------------------------ 7

Substitute eqn. 7 in 4,

[ (Rf + s Lf) (Js2 + Bs) (s) / Kf ] = Vf(s)

(s) (Rf + s Lf ) (Js2 + Bs) / Kf = Vf(s)

(s) Kf = Vf(s) (Rf + s Lf ) (Js2 + Bs)

Procedure:

(a) To find Rf :1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the field terminals of the trainer kit to the motor field terminal.5. Switch ON the power switch and SPST switch S1.6. Shaft should not rotate, when applying DC input voltage.7. Note the field voltage and current for the various values of field voltage. 8. Calculate field resistance Rf.

(b) To find Lf :1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the AC voltage terminal of the trainer kit to the motor field terminal.5. Switch ON the power.6. Shaft should not rotate, when applying AC input voltage.7. Note the AC voltage and current for the various values of AC voltage. 8. Calculate field inductance Lf.

(c) To find Kf :1. Initially keep all the switches in OFF position.2. Initially keep AC voltage adjustment POT in minimum position.3. Initially keep armature and field voltage adjustment POT in minimum position.4. Connect the armature terminals of the trainer kit to the motor armature terminal.5. Connect the field terminals of the trainer kit to the motor field terminal.6. Switch ON the power switch and SPST switches S1 and S2.7. Set the armature voltage at rated value (48 V).8. Vary field voltage till motor runs at rated speed.

9. By varying the load, note the field voltage, field current and spring balance readings.

10. Plot the graph T versus field current. From the graph calculate the value of Kf.

Tabulation:

To find Rf :

Average Rf = _________ Ω

To find Lf :

Field voltage

( Vf)

Field current

( If)

Field Impedence Zf = Vf / If

Armature ReactanceXf = ( Zf

2 – Rf2)

Armature InductanceLf = Xf / 2 f

Average Lf = _________ H

To find Kf :

Field voltage

( Vf)

Field current( If)

S1

(Kg)S2

(Kg)S1 - S2

(Kg)Torque

(T) = 9.81(S1-S2) R Nm

Field voltage( Vf)

Field current( If)

Field ResistanceRf = Vf / If

Model Graph:

Measurement of Kf :

Torque Vs Field current(If) :

T(Nm) T

If

If (A)

Model calculation :

Kf = T/ If

B = 0.001 Nm/ (rad/sec)J = 0.0074 Kg m2 R = 0.075 m

Result:Thus the transfer function of the field controlled DC servomotor is determined as

Ex. No. – 2

DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF AC SERVOMOTOR

Aim:To determine the transfer function parameters of AC servomotor.

Apparatus Required:

S.No. Apparatus Quantity1. AC servomotor trainer kit 1

2. AC servomotor 1

Name plate details of AC servomotor:

Main winding voltage 230 VControl winding voltage 230 VLoad current per phase 350 mAInput power 100 WFull load Speed 900 rpm

Theory:

It is an electromechanical device, which converts electrical signal into angular displacement. An AC servomotor basically a two-phase induction motor except for certain special design features. The rotor of the servomotor is built with high resistance, so that X / R ratio is small which result in linear speed-torque characteristics.

The stator consists of two pole pairs mounted on the inner periphery of the stator, such that their axes are at an angle of 90o in space. Each pole pairs carries a winding. One winding is called Reference winding and the other is called Control winding.

The rotor construction is usually squirrel cage .The squirrel cage rotor is made of laminations. The rotor bars are placed on the slots and short-circuited at both ends by end rings. The diameter of the rotor is small in order to reduce inertia and to obtain good accelerating characteristics. The Drag cup construction is employed for very low inertia application. In this type of construction the rotor will be in the form of hollow cylinder made of aluminum. The aluminum cylinder itself acts as short-circuited rotor conductors.

Front Panel diagram of AC servomotor:

Fig. 2.1 Circuit diagram of AC servomotor

Derivation of transfer function:

Let,Tm = Torque developed by servomotorQ = Angular displacement of rotorω = Angular speedTL = Torque required by the loadJ = Moment of inertiaB = Viscous frictional coefficientK1 = Slope of control-phase voltage Vs torque characteristicK2 = Slope of speed-torque characteristic

Torque developed by motor, Tm = K1 ec – K2 d / dt ----------------- 1

Load torque,

TL = J d2 / dt + B d / dt ------------------ 2

At equilibrium motor torque is equal to load torque

J d2 / dt + B d / dt = K1ec – K2 d / dt ------------------ 3 On taking laplace transform of above equation, we get

Js2 (s) + Bs (s) = K1 Ec(s) – K2 s (s) ( Js2+ Bs+K2 s) (s) = K1 Ec(s) (s) K1

= Ec(s) s (Js+B+K2)

(s) K1 / (B+K2 )

= Ec(s) s [ ( J s / (B+K2)) + 1]

(s) Km

= Ec(s) s (m s + 1)

Where, Km = K1 / B+K2 = Motor gain constant m = J / B+K2 = Motor time constant

Procedure:

To find K1 :1. Apply 3-phase AC supply to the 3-phase input terminal of the trainer kit.2. Switch on the power switch.3. Switch on the control winding and main winding switches S1 and S2.4. Apply rated voltage (230 V) to the reference phase winding.5. Apply a certain voltage to the control phase winding and make the motor run at

low speed.6. Apply load to motor. Motor speed will decrease. Increase the control voltage until

the motor runs at same speed as on no-load.7. Note down control voltage and load.8. Repeat steps 6 & 7 for various loads.9. Calculate the torque using the formula, T = 9.81 x Radius of shaft x Load. 10. Plot the graph between T versus control voltage.11. From the graph find the motor constant K1.

To find K2 :1. Apply 3-phase AC supply to the 3-phase input terminal of the trainer kit.2. Switch on the power switch.3. Switch on the control winding and main winding switches S1 and S2.4. Apply rated voltage to the reference phase winding and control phase winding.5. Apply the load in step by step upto the motor run at zero rpm and note the speed

of motor and applied load.6. Calculate the torque using the formula, T = 9.81 x Radius of shaft x Load. 7. Plot the graph between T versus Speed.8. From the graph find the motor constant K2.

Tabulation:

To find K1 :

S.No Control voltage(V)

Load(Kg)

Torque(Nm)

To find K2 :

S.NoSpeed (rpm)

Load(Kg)

Torque(Nm)

Model graph:

Torque Vs control voltage:

Torque (Nm)

Slope K1

Control voltage (V)

Torque Vs Speed:

Torque

Slope K2

Speed

Model calculation:

K1 = ΔT / ΔVK2 = ΔT / ΔN

T = 9.81 x Radius of shaft x Load

J = 0.0155 Kg m2

B = 0.85 x 10-4 kg m s

Result :Thus the transfer function parameters of AC servomotor is found and the transfer function is determined as,

Ex. No. – 3

ANALOG SIMULATION OF TYPE-0 AND TYPE-1 SYSTEM

Aim:To study the response of type-0 and type-1 system in open and closed loop.

Apparatus Required:

S.No. Apparatus Quantity1. Process simulator kit 1

2. CRO 1

3. Patch wires As required

Theory:

When a number of elements or components are connected in a sequence to perform a specific function, the group thus formed is called a SYSTEM. The input and output relationship of the system can be expressed by a differential equation.

Order of the system:The order of the system is given by the order of the differential equation governing the system. The input output relationship can be expressed by transfer function also. Transfer function of a system is obtained by taking laplace transform of the differential equation governing the system and rearranging them as ratio of two polynomials in ‘s’. The order is given by the maximum power of ‘s’ in denominator polynomial Q(s).

T(s) = P(s) / Q(s)P(s) - Numerator polynomialQ(s) - Denominator polynomial

Q(s) =ao sn + a1 sn-1 + a2 sn-2 + …………. + an-1 s + an

If n=0, then system is Zero-Order system.If n=1, then system is First-Order system.If n=0, then system is Second-Order system.

Type Number of system:Type number is given by the number of poles lying at the origin.

If N=0, the system is a Type Zero system.If N=1, the system is a Type One system.If N=0, the system is a Type Two system.

Block diagram:

Type-0 Open-loop system:

Type-0 closed-loop system:

Type-1 Open-loop system:

Type-1 closed-loop system:

Procedure:1. Give the connections as per the block diagram in the process simulator kit using

the front panel diagram.2. Set the input (set point) by using the set value knob.3. Observe the measured value or process value (PV) using CRO.4. Tabulate the readings and calculate the % error.5. Repeat the same for type-0 and type-1 first order systems in open and closed loop.6. Plot the response in the graph.

Front panel diagram:

Tabulation:

Type-0 first order system:

Set point(SP)

Process variable(PV)

Settling time Error =SP – PV

% error =(SP – PV) / PV

X 100Open loop

Closed loop

Type-1 first order system:

Set point(SP)

Process variable(PV)

Settling time Error =SP – PV

% error =(SP – PV) / PV

X 100Open loop

Closed loop

Model graph:

Response of first order system:

Result:Thus the analog simulation of type-0 and type-1 first order system has been observed using process control simulator.

Ex. No. – 4

DETERMINATION OF TRANSFER FUNCTION OF SEPERATELY EXCITED DC GENERATOR

Aim:To determine the transfer function of separately excited DC generator.

Apparatus Required:

S.No. Apparatus Quantity

1. Voltmeter

2. Ammeter

3 Rheostat

4. Connecting wires

5.

6.

Name plate details:

Generator Motor

Fuse Rating:Fuse Rating = 125% of rated current

Derivation of transfer function:

Applying KVL to the field side,

ef = Rf if + Lf (dif / dt) ............. 1Applying KVL to the armature side,

eg = Ra ia + La (dia / dt) + RL ia ............. 2

VL = RL ia .............. 3

Also eg α if , So, eg = Kg if ............. 4

Taking Laplace transform of equation (1) we get

Ef (s) = Rf If(s) + s Lf If(s)Ef (s) = If (s) [Rf + sLf]If (s) = Ef (s) / [Rf + sLf] ................. 5

Taking Laplace transform of equation (2) we get

Eg (s) = Ra Ia(s) + sLa Ia(s) + RL Ia(s)Eg (s) = Ia(s) [Ra + sLa + RL] ............... 6

Taking Laplace transform of equations (3) and (4) we get

VL(s) = RL Ia( s) Therefore, Ia( s) = VL(s) / RL ................ 7

Eg(s) = Kg If(s) ................ 8

Substituting. equations (7) and (8) in equation (6) we get

Kg If(s) = [Ra + sLa + RL] [VL(s) / RL] ............... 9

Substituting the value of If (s) in the above equation we get

Kg Ef (s) / [Rf + sLf] = [Ra + sLa + RL] [ VL(s) / RL]

Hence transfer function,

VL(s) / Ef (s) = Kg RL / [ (Rf + sLf) (Ra + sLa + RL) ]

Procedure:

(a) To find Ra:1. Connections are given as per the circuit diagram.2. Apply DC voltage to the armature terminals.3. By varying the resistance, note the armature voltage and armature current.4. Calculate the armature resistance by using the formula, Ra = Va / Ia .

(b) To find La:1. Connections are given as per the circuit diagram.2. Apply AC voltage to the armature terminals.3. By varying the resistance, note the AC armature voltage and AC armature current.4. Calculate the armature impedance by using the formula, Za = Va / Ia .5. Now calculate the armature reactance, Xa and then the armature Inductance, La.

(c) To find Rf:1. Connections are given as per the circuit diagram.2. Apply DC voltage to the field terminals.

3. By varying the resistance, note the field voltage and field current.4. Calculate the field resistance by using the formula, Rf = Vf / If .

(d) To find Lf:1. Connections are given as per the circuit diagram.2. Apply AC voltage to the field terminals.3. By varying the resistance, note the AC field voltage and AC field current.4. Calculate the armature impedance by using the formula, Zf = Vf / If .5. Now calculate the field reactance, Xf and then the field Inductance, Lf.

(e) To find Kg :1. Connections are made as shown in the circuit diagram.2. The motor field rheostat should be in minimum resistance position and the generator field rheostat should be in maximum resistance position while switching ON and switching OFF the supply side DPST switch.3. Ensure that the DPST switch on the load side is open.4. Switch ON the supply and DPST switch.5. The generator is brought to its rated voltage by varying the generator field rheostat. 6. The DPST switch on the load side is closed, and the load is varied for Convenient steps of load.7. Note load voltage VL and load current Ia . 8. Plot graph between VL and IL. The slope of the graph gives Kg.

Circuit diagram:

(a) To find Ra :

(b) To find La :

(c) To find Rf :

(d) To find Lf :

(e) To find Kg :

Tabulation:

To find Ra :

Average Ra = _________ Ω

Armature voltage( Va)

Armature current( Ia)

Armature ResistanceRa = Va / Ia

To find La :

Armature voltage

( Va)

Armature current

( Ia)

Armature Impedence Za=Va/Ia

Armature ReactanceXa= ( Za2 - Ra

2)

Armature InductanceLa= Xa/ 2 f

Average La = _________ H

To find Rf :

Average Rf = _________ ΩTo find Lf :

Field voltage

( Vf)

Field current

( If)

Field Impedence

Zf=Vf/If

Field ReactanceXf= ( Zf

2 – Rf2)

Field InductanceLf= Xf/ 2 f

Average Lf = _________ HTo find Kg :

Load voltage, VL

(V)Load current, IL

(A)

Model graph:

Field voltage( Vf)

Field current( If)

Field ResistanceRf = Vf / If

Result:Thus the transfer function of the separately excited DC generator is determined as,

Ex. No. – 5 (a)

DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC MOTOR

Aim:To determine the transfer function of armature controlled DC motor.

Apparatus Required:

S.No. Apparatus Quantity

1. Voltmeter

2. Ammeter

3 Rheostat

4. Connecting wires

5.

6.

Name plate details:

Motor

Fuse Rating:Fuse Rating = 125% of rated current

Theory:The DC motor converts electrical energy into mechanical energy. The electrical

energy supplied at the armature terminals is converted into controlled mechanical energy.

In armature control, the field current is kept constant and the armature voltage is varied and hence the speed is varied. The field current If is maintained constant by keeping the Vf constant and the armature current Ia is varied to change the torque T of the load connected to the motor shaft. Thus the input variable of the motor is the armature voltage Va and the output variable is the torque T. The speed of the DC motor is directly proportional to the armature voltage and inversely proportional to the flux in the armature. In the armature controlled DC motor, the desired speed is obtained by varying the armature voltage.

Derivation of transfer function:Let, Ra = Armature Resistance La = Armature Inductance Eb = Back emf Ia = Armature current T = Torque developed in the motor J = Moment of inertia B = Dashpot

= Angular DisplacementKt = Armature torque constantKb = Back emf constant

From equivalent circuit, Differential equation of electrical circuit can be written as Ra ia +La dia/dt + eb = Va ------------------------ 1

Differential equation of mechanical system can be written as

J d2 /dt2 + B d / dt = T ------------------------ 2

Torque is directly proportional to the armature current T Ia , T Ia T = Kt Ia ------------------------ 3

Motor back emf is directly proportional to the speed (Angular velocity ) Eb = Kb d / dt ------------------------ 4

Taking laplace transform for equations 1, 2, 3 & 4 we get

Ra Ia(s) + Las Ia(s) + Eb(s) = Va(s) (Ra + Las) Ia(s) + Eb(s) = Va(s) ------------------------ 5

Js2(s) + Bs(s) = T(s) ------------------------ 6

T(s) = Kt Ia(s) ------------------------ 7

Eb(s) = Kbs (s) ------------------------ 8

Equating equation 6 & 7 we get, Kt Ia(s) = Js2(s) + Bs(s)

Kt Ia(s) = (Js2 + Bs) (s)

Ia(s) = (Js2 + Bs) (s) / Kt ------------------------ 9

Substitute eqn. 8 & 9 in 5,

[ (Ra + Las) (Js2 + Bs) (s) / Kt ] + Kbs (s) = Va(s)

(s) (Ra + Las) (Js2 + Bs) + Kb Kt s / Kt = Va(s)

(s) Kt = Va(s) (Ra + Las) (Js2 + Bs) + Kb Kt s

Procedure:

(a) To find Ra:

5. Connections are given as per the circuit diagram.6. Apply DC voltage to the armature terminals.7. By varying the resistance, note the armature voltage and armature current.8. Calculate the armature resistance by using the formula, Ra = Va / Ia .

(b) To find La:

6. Connections are given as per the circuit diagram.7. Apply AC voltage to the armature terminals.8. By varying the resistance, note the AC armature voltage and AC armature current.9. Calculate the armature impedance by using the formula, Za = Va / Ia .10. Now calculate the armature reactance, Xa and then the armature Inductance, La.

(c) To find Kb:

1. Connections are given as per the circuit diagram.2. Keep the field rheostat in minimum position and switch on the supply.3. Adjust the field rheostat to rated speed.4. For various values of armature voltage Va, note Va, Ia and N.5. Plot the graph Eb versus ω. From the graph calculate the value of Kb.

(d) To find Kt:

1. Connections are given as per the circuit diagram.2. Set the field voltage at rated value .3. Vary armature voltage till motor runs at rated speed.4. By varying the load, note the armature voltage, current and spring balance

readings.5. Plot the graph T versus armature current. From the graph calculate the value of Kt.

Circuit diagram:

(f) To find Ra :

(g) To find La :

(h) To find Kb and Kt :

Tabulation:

To find Ra :

Average Ra = _________ ΩTo find La :

Armature voltage

( Va)

Armature current

( Ia)

Armature Impedence Za=Va/Ia

Armature ReactanceXa= ( Za2 - Ra

2)

Armature InductanceLa= Xa/ 2 f

Average La = _________ HTo find Kb :

Armature voltage

( Va)

Armature current

( Ia)

Speed( N )rpm

Eb =Va - IaRa

Volts =2N/60

rad/sec

To find Kt :

Armature current

( Ia)

Armature voltage

( Va)

S1

(Kg)S2

(Kg)S1 - S2

(Kg)Torque

(Ta) = 9.81(S1-S2) R Nm

Model Graph:

Armature voltage( Va)

Armature current( Ia)

Armature ResistanceRa = Va / Ia

Measurement of Back Emf constant (Kb) :

Eb

Eb

(radians/sec)

Measurement of Kt :

Torque Vs Armature current (Ia)

T(Nm) T

Ia

Ia (A)

Model calculation :

Kb = Eb /

Kt = T/ Ia

B = 0.001 Nm/ (rad/sec)J =0.0074 Kg m2

R = 0.075 m

Result:Thus the transfer function of the armature controlled DC motor is determined as

Ex. No. – 5 (b)

DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC MOTOR

Aim:To determine the transfer function of field controlled DC motor.

Apparatus Required:

S.No. Apparatus Quantity

1. Voltmeter

2. Ammeter

3 Rheostat

4. Connecting wires

5.

6.

Name plate details:

Motor

Fuse Rating:Fuse Rating = 125% of rated current

Theory:The DC motor converts electrical energy into mechanical energy. The electrical

energy supplied at the armature terminals is converted into controlled mechanical energy.

In field control method, the armature current Ia is maintained constant while the field voltage Vf is varied to control the speed or torque of the motor. Thus the input of the motor is field voltage Vf and the output is the motor speed, and the load displacement

Derivation of transfer function:Let,

Rf = field Resistance Lf = field Inductance T = Torque developed in the motor J = Moment of inertia B = Dashpot

= Angular DisplacementKf = Field torque constant

From equivalent circuit,

Differential equation of electrical circuit can be written as Rf if +Lf dif/dt = Vf ------------------------ 1

Differential equation of mechanical system can be written as

J d2 /dt2 + B d / dt = T ------------------------ 2

Torque is directly proportional to the armature current T Ia , T If T = Kf If ------------------------ 3

Taking laplace transform for equations 1, 2, & 3 we get

Rf If(s) + s Lf If(s) = Vf(s) (Rf + s Lf) If(s) = Vf(s) ------------------------ 4

Js2(s) + Bs(s) = T(s) ------------------------ 5

T(s) = Kf If(s) ------------------------ 6

Equating equation 5 & 6 we get, Kf If(s) = Js2(s) + Bs(s)

Kf If(s) = (Js2 + Bs) (s)

If(s) = (Js2 + Bs) (s) / Kf ------------------------ 7

Substitute eqn. 7 in 4,

[ (Rf + s Lf) (Js2 + Bs) (s) / Kf ] = Vf(s)

(s) (Rf + s Lf ) (Js2 + Bs) / Kf = Vf(s)

(s) Kf = Vf(s) (Rf + s Lf ) (Js2 + Bs)

Procedure:

(d) To find Rf:

5. Connections are given as per the circuit diagram.6. Apply DC voltage to the field terminals.7. By varying the resistance, note the field voltage and field current.8. Calculate the field resistance by using the formula, Rf = Vf / If .

(e) To find Lf:

7. Connections are given as per the circuit diagram.8. Apply AC voltage to the field terminals.9. By varying the resistance, note the AC field voltage and AC field current.10. Calculate the armature impedance by using the formula, Zf = Vf / If .11. Now calculate the field reactance, Xf and then the field Inductance, Lf.

(f) To find Kf:

1. Connections are given as per the circuit diagram.2. Set the field voltage at 50 % rated value.3. Set the armature voltage at 50 % rated value.4. By varying the load, note the field voltage, current and spring balance

readings.5. Plot the graph T versus field current. From the graph calculate the value of Kf.

Circuit diagram:

(a) To find Rf :

(b) To find Lf :

(c) To find Kf :

Tabulation:

To find Rf :

Average Rf = _________ Ω

Field voltage( Vf)

Field current( If)

Field ResistanceRf = Vf / If

To find Lf :

Field voltage

( Vf)

Field current

( If)

Field Impedence

Zf=Vf/If

Field ReactanceXf= ( Zf

2 – Rf2)

Field InductanceLf= Xf/ 2 f

Average Lf = _________ HTo find Kf :

Field current

( If)

Field voltage ( Vf)

S1

(Kg)S2

(Kg)S1 - S2

(Kg)Torque

(T) = 9.81(S1-S2) R Nm

Model Graph:

Measurement of Kf :

Torque Vs Field current (If)

T(Nm) T

If

If (A)

Model calculation :

Kf = T/ If

B = 0.001 Nm/ (rad/sec)J =0.0074 Kg m2

R = 0.075 m

Result:Thus the transfer function of the field controlled DC motor is determined as

Ex. No. – 6

DIGITAL SIMULATION OF FIRST-ORDER SYSTEMSAim:

To study the time response characteristics of the first-order system.

Software used:Matlab

Simulation of First-order system:

Open loop system:

1. For step input:

Code: num=[1];den=[4 2];sys=tf(num,den);step(sys)

2. For impulse input:

Code: num=[1];den=[4 2];sys=tf(num,den);impulse(sys)

3. For sine input:

closed loop system:

1. For step input:

Code:num=[1];den=[1 1];sys1=tf(num,den);sys2=1;sys=feedback(sys1,sys2);step(sys)

2. For impulse input:

Code:num=[1];den=[1 1];sys1=tf(num,den);sys2=1;sys=feedback(sys1,sys2);impulse(sys)

3. For sine input:

Result:Thus the response of first-order system has been simulated and studied.